Archived posting to the Leica Users Group, 2011/11/03
[Author Prev] [Author Next] [Thread Prev] [Thread Next] [Author Index] [Topic Index] [Home] [Search]For sharper images, use the 55mm 1.8. I have them both. I find that sharpness trumps speed and bokeh in producing images that I am happy with. Jim Nichols Tullahoma, TN USA ----- Original Message ----- From: "philippe.amard" <philippe.amard at sfr.fr> To: "Leica Users Group" <lug at leica-users.org> Sent: Thursday, November 03, 2011 4:07 PM Subject: Re: [Leica] 1.8 vs. 1.4!?!? > Now some manufacturers - Pentax to name just one - had 50mm 1.4 and 55mm > 1.8 on their catalogues: which of these offered better bokeh or faster > takes? > > I'm so happy to have a full auto mode on my camera right now. > > Philippe looking for aspirin and about to call Doctor Ted > > > > Le 3 nov. 11 ? 21:40, Herbert Kanner a ?crit : > >> After this, the dead horse should stay dead. >> >> While your mathematics is correct, I differ with your conclusions. It >> relates to a distinction between "f numbers" and "stops". Agreed: >> reciprocal of the square of the f number is proportional to the amount >> of light getting through. But the question was: "What fraction of a stop >> is the difference between two f numbers?". >> >> Now, the stops on a lens aperture ring are an arbitrary choice, >> general;y meaning a factor of two in exposure. So going, say three stops >> toward smaller f numbers doubles the exposure three times, or 2 to the >> power 3. Now, let's think about this example of three stops: f/4, f/5.6, >> and f/11, pretend we don't know that those three numbers are engraved on >> the aperture ring, and ask: "How many stops are the distance between f/4 >> and f/11. First, realize that the conventional set of stops aren't >> exactly a factor 2 in exposure because the lens manufacturers in their >> wisdom wanted the f numbers to not have more than two digits; who would >> like 3.99762 engraved on their aperture ring? So, f/4 to f/11 is >> actually a change in exposure of 7.5625, close enough to 8. >> >> Next let's go backward, again pretending ignorance of aperture ring >> numbers, and ask how many "stops" change is f/4 to f/11. take 11/4 and >> square it, getting, as before, 7.5625. Take the logarithm (base 2) of >> 7.5625, and you get 2.918, which is the actual theoretical number of >> stops, and close enough to the three notches on the ring. >> >> That is the way I calculated the number of stops between 1.8 and 1.4. >> >> How do you calculate log(base 2) of something. Just divide its log(base >> 10) by log of 2 (base 10) >> >> Herb >> >> >> >>> Time to beat a dead horse! If anyone is interested: >>> >>> It is all geometry, namely the area of a circle. >>> >>> For each f-stop, we have double the light. The f-stop is related to >>> the size of the aperture, which is approximated as a circle. The >>> amount of light coming is proportional to the circle's area, which >>> you may recall is pi times the radius squared, pi*r^2. We use the >>> f# for the equivalent radius. >>> >>> Thus, starting with f1, and r = 1, the area is pi*r^2 = pi*1*1 = pi, >>> as the relative amount of light. >>> >>> For an amount of light 2*pi (next f-stop, double the light), pi*r^2 >>> = 2 pi. Divide both sides by pi, and you get r^2 =2. r = the >>> square root of 2, or 1.414? f1.4 is the next stop. >>> >>> This is where the 1.4 factor George mentioned comes from; the square >>> root of 2 is 1.414. >>> >>> Next f-stop, double the light again: pi*r^2 = 2*2 pi. r^2 = 4, f2 >>> is the next stop. >>> >>> So, if you want fractional stops: >>> >>> 1/3 stop: Square root of 1.333 = 1.15456 >>> 1/2 stop: Square root of 1.5 = 1.22474 >>> 2/3 stop: Square root of 1.667 = 1.29112 >>> >>> So going back to Mark's f1.4 example: >>> 1/3 stop slower = 1.414 * 1.15456 = f1.633 or f1.6 >>> 1/2 stop slower = 1.414 * 1.22474 = f1.732 or f 1.7 >>> 2/3 stop slower = 1.414 * 1.29112 = f1.828 or f1.8 >>> 1 stop slower = 1.414 * 1.414 = f2.0 >>> >>> >>> Matt >>> >>> >>> >>> >>> >>> >>> >>> On Nov 2, 2011, at 5:32 PM, Mark Rabiner wrote: >>> >>> > I looked up f 1.8 vs. 1.4 thinking it was between a half and a >>> quarter of a >>> > stop and they are saying its 2/3rds!?!?! Anybody know that that's >>> true? >>> > >>> > Where is there a photo calculator that tells you these things?!?!? >>> >>> I always thought the basic math for 1 f stop revolved around a factor >>> of 1.4. >>> 1.4 x 1.4 = 1.96 >>> 1.8 / 1.4 = 1.29 >>> >>> so - yes - 2/3 would seem close enough for? >>> what? I'm not sure. >>> >>> Regards, >>> George Lottermoser >>> george at imagist.com >>> http://www.imagist.com >>> http://www.imagist.com/blog >>> http://www.linkedin.com/in/imagist >>> >>> >>> >>> _______________________________________________ >>> Leica Users Group. >>> See http://leica-users.org/mailman/listinfo/lug for more information >> >> -- >> Herbert Kanner >> kanner at acm.org >> 650-326-8204 >> >> Question authority and the authorities will question you. >> >> _______________________________________________ >> Leica Users Group. >> See http://leica-users.org/mailman/listinfo/lug for more information > > _______________________________________________ > Leica Users Group. > See http://leica-users.org/mailman/listinfo/lug for more information > >